Question

Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:

• A. \( 3a_{99} + 100 \)
• B. \( 3a_{99} - 100 \)
• C. \( 3a_{100} + 100 \)
• D. \( 3a_{100} - 100 \)

Hint:

Utilize characteristic equations to solve linear recurrence relations efficiently.

Answer 1

The Correct Option is C

To determine \( \sum_{k=1}^{100} a_k \), we must first find the explicit form of \( a_n \) using the given recurrence relation and initial conditions. The sequence is defined by \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \). We assume a solution of the form \( a_n = r^n \). Substituting this into the recurrence yields \( 2r^{n+2} = 5r^{n+1} - 3r^n \). Dividing by \( r^n \) (assuming \( r eq 0 \)) gives the characteristic equation: \( 2r^2 = 5r - 3 \), which rearranges to \( 2r^2 - 5r + 3 = 0 \). Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2 \), \( b = -5 \), \( c = 3 \), we find the roots: \[ r = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4} \] The distinct roots are \( r_1 = \frac{6}{4} = \frac{3}{2} \) and \( r_2 = \frac{4}{4} = 1 \). The general solution is therefore \( a_n = A \left(\frac{3}{2}\right)^n + B(1)^n \). Applying the initial conditions \( a_0 = 0 \) and \( a_1 = \frac{1}{2} \): \[ a_0 = A(1)^0 + B(1)^0 = A + B = 0 \] \[ a_1 = A \left(\frac{3}{2}\right)^1 + B(1)^1 = \frac{3}{2}A + B = \frac{1}{2} \] From \( A + B = 0 \), we have \( B = -A \). Substituting this into the second equation: \[ \frac{3}{2}A - A = \frac{1}{2} \Rightarrow \frac{1}{2}A = \frac{1}{2} \Rightarrow A = 1 \] Consequently, \( B = -1 \). The explicit form of \( a_n \) is \( a_n = \left(\frac{3}{2}\right)^n - 1 \). Now we can compute the sum: \[ \sum_{k=1}^{100} a_k = \sum_{k=1}^{100} \left(\left(\frac{3}{2}\right)^k - 1\right) = \sum_{k=1}^{100}\left(\frac{3}{2}\right)^k - \sum_{k=1}^{100} 1 \] The first part is the sum of a geometric series with first term \( a = \frac{3}{2} \) and common ratio \( r = \frac{3}{2} \). The sum of the first 100 terms is \( S_{100} = \frac{\frac{3}{2}\left(\left(\frac{3}{2}\right)^{100} - 1\right)}{\frac{3}{2} - 1} = \frac{\frac{3}{2}\left(\left(\frac{3}{2}\right)^{100} - 1\right)}{\frac{1}{2}} = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right) \). The second part is \( \sum_{k=1}^{100} 1 = 100 \). Therefore, \[ \sum_{k=1}^{100} a_k = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right) - 100 \] Note that \( \left(\frac{3}{2}\right)^{100} - 1 = a_{100} \). So the sum is \( 3a_{100} - 3 - 100 \). There appears to be a slight discrepancy in the derivation presented in the prompt. Revisiting the sum calculation: \[ \sum_{k=1}^{100} a_k = \sum_{k=1}^{100} \left(\left(\frac{3}{2}\right)^k - 1\right) = \sum_{k=1}^{100} \left(\frac{3}{2}\right)^k - \sum_{k=1}^{100} 1 \] The geometric series sum from \( k=1 \) to \( 100 \) is \( \frac{\frac{3}{2}((\frac{3}{2})^{100}-1)}{\frac{3}{2}-1} = \frac{\frac{3}{2}((\frac{3}{2})^{100}-1)}{\frac{1}{2}} = 3((\frac{3}{2})^{100}-1) \). Then \( \sum_{k=1}^{100} a_k = 3((\frac{3}{2})^{100}-1) - 100 \). The prompt states \( \sum_{k=1}^{100} a_k = 3a_{100} + 100 \), which does not match the derived formula. If \( a_n = (\frac{3}{2})^n - 1 \), then \( \sum_{k=1}^{100} a_k = 3((\frac{3}{2})^{100}-1) - 100 = 3(\frac{3}{2})^{100} - 3 - 100 = 3(\frac{3}{2})^{100} - 103 \). The prompt's final statement is \( 3a_{100} + 100 \). Let's re-examine the sum calculation presented in the prompt: \( \sum_{k=1}^{100} a_k = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right) \). This seems to be the sum of the geometric part only. The prompt then incorrectly states \( \sum_{k=1}^{100} a_k = 3a_{100} + 100 \). There seems to be an error in the provided summation result. The correct sum of \( a_k = (\frac{3}{2})^k - 1 \) from \( k=1 \) to \( 100 \) is \( 3((\frac{3}{2})^{100}-1) - 100 = 3(\frac{3}{2})^{100} - 3 - 100 = 3(\frac{3}{2})^{100} - 103 \). This does not match \( 3a_{100} + 100 \).